Under the assumption that $F(2i,n) = 0$ when $i<4$ or $2i>n$, the first and second cases in the recurrence become partial cases of the third one. Hence, the recurrence reduces to
$$F(2i,n) =
\begin{cases} 
1, & \text{if } i=2,\ n=4;\\
0, & \text{if } i<2\text{ or } 2i>n;\\
\frac{2i+n-4}{2n-5}F(2i,n-1)+\frac{n-2i+1}{2n-5}F(2i-2,n-1) & \text{otherwise}.
\end{cases}
$$

Consider the generating function
$$f(x,y) := \sum_{i\geq 2} \sum_{n\geq 2i} F(2i,n) x^{n+2i-8} y^{n-2i+1}.$$

Then the first two cases imply $f(x,0) = 0$ and $f(0,y)=y$.

Now, the third case translates into a linear PDE:
$$\frac{\partial}{\partial x} (x^2\cdot f(x,y)) + xy\frac{\partial}{\partial y} f(x,y) = 3xy + \frac{y}{x^2}\frac{\partial}{\partial x} (x^5\cdot f(x,y)) + x^4y\frac{\partial}{\partial y} (y^{-1}\cdot f(x,y)),$$
which simplifies to
$$(x^2y^2-xy)f_x + (x^3y - y^2)f_y = (2y+x^3-5xy^2)f - 3y^2.$$